1. CSE351: Memory, Data, & Addressing I CSE 351 Spring 2017
Instructor:
Ruth Anderson
Teaching Assistants:
Dylan Johnson
Kevin Bi
Linxing Preston Jiang
Cody Ohlsen
Yufang Sun
Joshua Curtis

2. To-Do List Start-of-Course survey [Catalyst] due Thursday (3/30)
Section 1 is this Thursday
Install the virtual machine (VM) before coming to section
Bring your computer with you to section
Lab 0, due Monday 11:59pm
Homework 1, due Monday 11:59pm
Readings in CSAPP

3. Lab 0, Homework 1, and Readings
Lab 0, due Monday 11:59pm
Basic exercises to start getting familiar with C – need the VM
Credit/no-credit
Do ASAP, attending Section 1 will help
Homework 1, due Monday 11:59pm
3 canvas quizzes, 20 tries each, you best overall score is kept
Course policies – you can do this one now!
Unsigned Number Representations – will discuss on Wed
Number Bases - will discuss on Wed
Readings:
For Mon: CSPP: § (pp. 1-28)
For Wed: CSPP: § (pp )
For Fri: CSPP: § (pp )

4. Roadmap C: Java: Assembly language: OS: Machine code: Computer system:
Memory & data
Integers & floats
x86 assembly
Procedures & stacks
Executables
Arrays & structs
Memory & caches
Processes
Virtual memory
Memory allocation
Java vs. C C:
Java:
car *c = malloc(sizeof(car));
c->miles = 100;
c->gals = 17;
float mpg = get_mpg(c);
free(c);
Car c = new Car();
c.setMiles(100);
c.setGals(17);
mpg = c.getMPG();
Assembly language:
get_mpg:
pushq %rbp
movq %rsp, %rbp
...
popq %rbp
ret
OS:
Machine code:
Computer system:

5. Hardware: Logical View
Memory
CPU
Bus
CPU: all computations done here
Memory: big array of bytes
Big pipe between CPU and memory, in order to keep CPU fed
Bus: connect everything else to each other
Bus just means you broadcast on it (to everyone)
Same root as the other kind of “bus”
Disks bigger storage, slower, farther away, persistent
Network, etc.
Disks
USB
Etc.
Net

6. Hardware: Physical View
USB…
Bus connections
CPU
(empty slot)
CPU
Looks funny because of heat sink
I/O
controller
Memory
Storage connections

7. Hardware: 351 View (version 0)
Memory
instructions ?
CPU
data
CPU executes instructions; memory stores data
To execute an instruction, the CPU must:
fetch an instruction;
fetch the data used by the instruction; and, finally,
execute the instruction on the data…
which may result in writing data back to memory
The CPU and Memory are what this course focuses on.

8. Hardware: 351 View (version 1)
Memory
instructions
i-cache
take 470…
this week…
CPU
data
registers
The CPU holds instructions temporarily in the instruction cache
The CPU holds data temporarily in a fixed number of registers
Instruction and operand fetching is hardware-controlled
Data movement is programmer-controlled (in assembly)
We’ll learn about the instructions the CPU executes – take CSE/EE470 to find out how it actually executes them

9. Hardware: 351 View (version 1)
Memory
instructions
i-cache
take 470…
this week…
How are data and instructions represented?
CPU
data
registers
The CPU holds instructions temporarily in the instruction cache
The CPU holds data temporarily in a fixed number of registers
Instruction and operand fetching is hardware-controlled
Data movement is programmer-controlled (in assembly)
We’ll learn about the instructions the CPU executes – take CSE/EE470 to find out how it actually executes them
How does a program find its data in memory?
We’re going to start to answer these two questions today

10. Memory, Data, and Addressing
Representing information as bits and bytes
Organizing and addressing data in memory
Manipulating data in memory using C

11. Question 1: Memory CPU instructions data i-cache
take 470…
this week…
How are data and instructions represented?
CPU
data
registers
Okay so first…

12. Binary Representations
Base 2 number representation
A base 2 digit (0 or 1) is called a bit
Represent as or
Electronic implementation
Easy to store with bi-stable elements
Reliably transmitted on noisy and inaccurate wires
Leading zeros
0.0V
0.5V
2.8V
3.3V 1
Leading zeros…
Speed of transition from low to high and back limits the speed over our computers.
Stay away from middle voltages.
How fast we can get there is determined by… physics!

13. Review: Number Bases Key terminology: digit (d) and base (B)
In base B, each digit is one of B possible symbols
Value of 𝑖-th digit is 𝑑× 𝐵 𝑖 where 𝑖 starts at 0 and increases from right to left
n digit number dn-1dn d1d0
value = dn-1Bn-1 + dn-2Bn d1B1 + d0B0
In a fixed-width representation, left-most digit is called the most-significant and the right-most digit is called the least-significant
Notation: Base is indicated using either a prefix or a subscript 13

14. Describing Byte Values
0000 1
0001 2
0010 3
0011 4
0100 5
0101 6
0110 7
0111 8
1000 9
1001 A 10
1010 B 11
1011 C 12
1100 D 13
1101 E 14
1110 F 15
1111
Hex
Decimal
Binary
Binary ( – )
Byte = 8 bits (binary digits)
Decimal (010 – )
Hexadecimal (0016 – FF16)
Byte = 2 hexadecimal (or “hex” or base 16) digits
Base 16 number representation
Use characters ‘0’ to ‘9’ and ‘A’ to ‘F’
Write FA1D37B16 in the C language
as 0xFA1D37B or 0xfa1d37b
More on specific data types later…
LSB
MSB 1
0*27
0*26
1*25
0*24
1*23
1*22
0*21
1*20 32 8 4
= 4510
How familiar are you all with binary?
Upper/lower case is fine

15. Question 2: Memory CPU instructions data i-cache registers
take 470…
this week…
CPU
data
registers
How does a program find its data in memory?

16. Byte-Oriented Memory Organization
00•••0
FF•••F
• • •
Conceptually, memory is a single, large array of bytes, each with a unique address (index)
The value of each byte in memory can be read and written
Programs refer to bytes in memory by their addresses
Domain of possible addresses = address space
But not all values (e.g., 351) fit in a single byte…
Store addresses to “remember” where other data is in memory
How much memory can we address with 1-byte (8-bit) addresses?
Many operations actually use multi-byte values
Each byte can be read and written
Address space determined by how many bits in the address
So what if things don’t fit in a single byte?
How many things does 1 byte allow us to address?

17. Machine Words Word size = address size = register size
Word size bounds the size of the address space and memory
word size = 𝑤 bits → 2𝑤 addresses
Current x86 systems use 64-bit (8-byte) words
Potential address space: 𝟐𝟔𝟒 addresses 264 bytes  1.8 x 1019 bytes = 18 billion billion bytes = 18 EB (exabytes) = 16 EiB (exbibytes)
Actual physical address space: 48 bits
Instead, we typically talk about addresses in terms of “words”
Use same word size anywhere that you typically store addresses
You could count the world population with 33 fingers (using population number of billion)
[x86-64 currently supports up to 48-bit physical addresses: 256 TB (terabytes)]
How big is an exabyte?

18. Aside: Units and Prefixes
Here focusing on large numbers (exponents > 0)
Note that 103 ≈ 210
SI prefixes are ambiguous if base 10 or 2
IEC prefixes are unambiguously base 2
Disk capacity SI to IEC conversion

19. Word-Oriented Memory Organization
64-bit
Words
32-bit
Words
Addr.
(hex)
Bytes
Addresses specify locations of bytes in memory
Address of word = address of first byte in word
Addresses of successive words differ by word size (in bytes): e.g., 4 (32-bit) or 8 (64-bit)
Address of word 0, 1, … 10?
0x00
Addr = ??
0x01
0x02
Addr = ??
0x03
0x04
Addr = ??
0x05
0x06
0x07
0x08
Addr = ??
0x09
What do we use to address bigger chunks of bytes?
0x0A
Addr = ??
0x0B
0x0C
Addr = ??
0x0D
0x0E
0x0F

20. Word-Oriented Memory Organization
64-bit
Words
32-bit
Words
Addr.
(hex)
Bytes
Addresses still specify locations of bytes in memory
Address of word = address of first byte in word
Addresses of successive words differ by word size (in bytes): e.g., 4 (32-bit) or 8 (64-bit)
Address of word 0, 1, … 10?
Alignment
0x00
Addr = ??
0x01
0x02
0000
0004
0008
0012
Addr = ??
0x03
0x04
0000
0008
Addr = ??
0x05
0x06
0x07
0x08
Addr = ??
0x09
Use the first byte of the word
Still count by bytes
This means there will by a bunch of addresses we should never see
Valid addresses are correctly aligned
Aside: isn’t this wasteful?
Yes, but remember how much we can address with 64 bits? Probably don’t need to worry too much.
0x0A
Addr = ??
0x0B
0x0C
Addr = ??
0x0D
0x0E
0x0F

21. A Picture of Memory (64-bit view)
A “64-bit (8-byte) word-aligned” view of memory:
In this type of picture, each row is composed of 8 bytes
Each cell is a byte
A 64-bit pointer will fit on one row
one word
0x04
0x05
0x06
0x07
0x00
0x01
0x02
0x03
Address
0x00 0x 0x 0x
In this course, we’ll be doing everything in 64-bits
What’s the second row’s address going to be?
0x08
Third?
0x10 0x 0x 0x 0x 0x 0x

22. A Picture of Memory (64-bit view)
A “64-bit (8-byte) word-aligned” view of memory:
In this type of picture, each row is composed of 8 bytes
Each cell is a byte
A 64-bit pointer will fit on one row
one word
0x04
0x05
0x06
0x07
0x00
0x01
0x02
0x03
Address
0x00
0x08
0x10
0x18
0x20
0x28
0x30
0x08
0x09
0x0A
0x0B
0x0C
0x0D
0x0E
0x0F
0x38
0x40
0x48

23. Addresses and Pointers
64-bit example
(pointers are 64-bits wide)
An address is a location in memory
A pointer is a data object that holds an address
Address can point to any data
Value 351 stored at address 0x08
35110 = 15F16 = 0x F
Pointer stored at 0x38 points to address 0x08
Address
0x00
0x08
0x10
0x18
0x20
0x28
0x30
0x38
0x40
0x48 00 01 5F
Note that everything is just padded with extra 0s 00 08

24. Addresses and Pointers
64-bit example
(pointers are 64-bits wide)
An address is a location in memory
A pointer is a data object that holds an address
Address can point to any data
Pointer stored at 0x48 points to address 0x38
Pointer to a pointer!
Is the data stored at 0x08 a pointer?
Could be, depending on how you use it
Address
0x00
0x08
0x10
0x18
0x20
0x28
0x30
0x38
0x40
0x48 00 01 5F
Note that everything is just padded with extra 0s 00 08 00 38

25. address size = word size
Data Representations
Sizes of data types (in bytes)
Java Data Type
C Data Type
32-bit (old)
x86-64
boolean
bool 1
byte
char 2
short
short int
int 4
float
long int 8
double
long
long double 16
(reference)
pointer *
This mapping is mostly an arbitrary historical artifact except for address sizes.
(reference)
pointer * 4 8
address size = word size
To use “bool” in C, you must #include <stdbool.h>

26. More on Memory Alignment in x86-64
For good memory system performance, Intel recommends data be aligned
However the x86-64 hardware will work correctly regardless of alignment of data
Design choice: x86-64 instructions are variable bytes long
Aligned: Primitive object of 𝐾 bytes must have an address that is a multiple of 𝐾
More about alignment later in the course 𝐾
Type 1
char 2
short 4
int, float 8
long, double, pointers

27. Byte Ordering How should bytes within a word be ordered in memory?
Example: store the 4-byte (32-bit) int: 0x a1 b2 c3 d4
By convention, ordering of bytes called endianness
The two options are big-endian and little-endian
Based on Gulliver’s Travels: tribes cut eggs on different sides (big, little)

28. Byte Ordering Big-endian (SPARC, z/Architecture)
Least significant byte has highest address
Little-endian (x86, x86-64)
Least significant byte has lowest address
Bi-endian (ARM, PowerPC)
Endianness can be specified as big or little
Example: 4-byte data 0xa1b2c3d4 at address 0x100
Big Endian
Most significant is first
Little Endian
Looks backwards
Let’s look at some more examples…
0x100
0x101
0x102
0x103 01 23 45 67
Big Endian a1 b2 c3 d4
0x100
0x101
0x102
0x103 67 45 23 01
Little Endian d4 c3 b2 a1

29. Byte Ordering Examples
Decimal: 12345
Binary:
Hex:
Byte Ordering Examples
IA32, x86-64 (little endian)
SPARC (big endian)
int x = 12345;
// or x = 0x3039;
0x00 39 30 00 30 39 00
0x00
0x01
0x01
0x02
0x02
0x03
0x03 30 39 00
32-bit
SPARC 30 39 00
64-bit
SPARC
long int y = 12345;
// or y = 0x3039;
(A long int is the size of a word) 39 30 00
IA32 39 30 00
x86-64
0x00
0x01
0x02
0x03
0x00
0x01
0x02
0x03
0x00
0x01
0x02
0x03 00
0x00
0x01
0x02
0x03 00
0x04
0x05
0x06
0x07
0x04
0x05
0x06
0x07

30. Endianness
Often programmer can ignore endianness because it is handled for you
Bytes wired into correct place when reading or storing from memory (hardware)
Compiler and assembler generate correct behavior (software)
Endianness still shows up:
Logical issues: accessing different amount of data than how you stored it (e.g. store int, access byte as a char)
When running down memory errors, need to know exact values
Manual translation to and from machine code (in 351)

31. Reading Byte-Reversed Listings
32-bit example
Disassembly
Take binary machine code and generate an assembly code version
Does the reverse of the assembler
Example instruction in memory
add value 0x12ab to register ‘ebx’ (a special location in the CPU)
Address Instruction Code Assembly Rendition
: 81 c3 ab add $0x12ab,%ebx
1-to-1 mapping between binary instructions and assembly instructions
Disassembler reverses the process of the assembler
You’ll see this in Lab 2, comes up when debugging binaries using GDB
$0x12ab represents a literal value (4779)
Next slide
Deciphering numbers

32. Reading Byte-Reversed Listings
32-bit example
Disassembly
Take binary machine code and generate an assembly code version
Does the reverse of the assembler
Example instruction in memory
add value 0x12ab to register ‘ebx’ (a special location in the CPU)
Address Instruction Code Assembly Rendition
: 81 c3 ab add $0x12ab,%ebx
Note how the literal value
Deciphering numbers
Value: 0x12ab
Pad to 32 bits: 0x000012ab
Split into bytes: ab
Reverse (little-endian): ab

33. Question:
We store the value 0x as a long int (size of long int = size of pointer) at address 0x100 and then get back 0x00 when we read a byte at address 0x102
What machine setup are we using?
32-bit, big-endian
(A)
32-bit, little-endian
(B)
64-bit, big-endian
(C)
64-bit, little-endian
(D)

34. Summary Memory is a long, byte-addressed array
Word size bounds the size of the address space and memory
Different data types use different number of bytes
Address of chunk of memory given by address of lowest byte in chunk
Object of 𝐾 bytes is aligned if it has an address that is a multiple of 𝐾
IEC prefixes refer to powers of 2 10
Pointers are data objects that holds addresses
Endianness determines storage order for multi-byte objects